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I am constructing the circuit given below: enter image description here

The problem, however, is that when I connect the electrodes to my skin, I get a mild shock. I am powering the Arduino using an USB cable. I connected the grounds in the image to the Arduino's ground. I am guessing that the USB power is causing problems. Any suggestions on how to properly ground the Arduino? Also, I am not using any batteries in this circuit. The only power source is the USB cable that I use to power the Arduino. The +5 volt go to the Arduino's 5V pin and the ground ones go to the ground pin on the Arduino.

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    \$\begingroup\$ Without proper isolation, you should not be connecting this circuit to your body while it is connected to a USB port. \$\endgroup\$
    – Matt Young
    Feb 17, 2016 at 16:06
  • \$\begingroup\$ How then, do i provide the isolation? Any ideas? \$\endgroup\$ Feb 17, 2016 at 16:10
  • \$\begingroup\$ Isolation is not hard to achieve, but it requires attention to detail and really is worth a question on its own. It's not something you're going to get using an Arduino board. \$\endgroup\$
    – Matt Young
    Feb 17, 2016 at 16:15
  • \$\begingroup\$ So then, should I just use the circuit as it is using resistors on the leads? \$\endgroup\$ Feb 17, 2016 at 16:19
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    \$\begingroup\$ The 500k on each lead needs to be there. To meet medical safety directives, it should technically be 2x249k on each. I would never hook this up to my skin without proper isolation. If you're going to use it, I can't stop you. You're the one taking the risk. \$\endgroup\$
    – Matt Young
    Feb 17, 2016 at 16:23

1 Answer 1

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Put 500kOhm resistors on each electrode lead.

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  • \$\begingroup\$ Thanks for the advice. But are my ground connections correct? I don't want to damage the board. \$\endgroup\$ Feb 17, 2016 at 16:06
  • \$\begingroup\$ @SwastikMohapatra Yes, assuming that it is built as it your drawing \$\endgroup\$ Feb 17, 2016 at 16:07
  • \$\begingroup\$ Just one more thing, I am using this to measure EEG signals. Won't a 500k resistor block those tiny signals? \$\endgroup\$ Feb 17, 2016 at 16:11
  • \$\begingroup\$ @SwastikMohapatra -- no, the input resistance of the AD620 is very high. I would suggest matching the resistors as closely as possible, or better yet, include the capacitors on the inputs that are shown in the data sheet for that integrated circuit. Also, you don't seem to have a gain setting resistor for the AD620, and you will need one. \$\endgroup\$ Feb 17, 2016 at 16:21
  • \$\begingroup\$ I know, the schematic is really horrible. I used a 2.2K resistor for an approximate gain of 23. So, according to you, I should add 2 capacitors as given in the data sheet? \$\endgroup\$ Feb 17, 2016 at 16:24

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